Prove without matrix expansion and only with row and column operations that
$\left| \begin{matrix}
   ah+bg & g & ab+ch \\
   bf+ba & f & hb+bc \\
   af+bc & c & bg+fc \\
\end{matrix} \right|=a\left| \begin{matrix}
   ah+bg & a & h \\
   bf+ba & h & b \\
   af+bc & g & f \\
\end{matrix} \right|$

Hint:This question is about matrices, therefore we need to use the properties of matrices and use them effectively to convert the left hand side of the equation to the form given in the right hand side. In particular, we have to use the property that the determinant remains the same if we add one row by some multiple of the other row and that if we multiply any row or column of the determinant by some number, the value of the determinant becomes multiplied by the cube of that number.

Complete step-by-step answer:
We need to show that the determinant value is the same in the left hand side and the right hand side. For. This we shall use two properties of the determinants, which are
(i) If we add one row by some multiple of the other row, the value of the determinant remains the same…………………………………..(1.1)
(ii) If any row or column of the determinant is multiplied by some number, the value of the determinant becomes multiplied by the cube of that number…………………………(1.2)
(iii) When, two determinants are added, their corresponding elements are added together.
Now, if we multiply c to column 2 and then multiply an overall $\dfrac{1}{c}$, the overall determinant remains the same from equation (1.1), thus
$\left| \begin{matrix}
   ah+bg & g & ab+ch \\
   bf+ba & f & hb+bc \\
   af+bc & c & bg+fc \\
\end{matrix} \right|=\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & gc & ab+ch \\
   bf+ba & fc & hb+bc \\
   af+bc & {{c}^{2}} & bg+fc \\
\end{matrix} \right|..........(1.3)$
Similarly, if we multiply c to third column and then add b times column 2 to column 3 in the determinant in the RHS, we get
$\begin{align}
  & a\left| \begin{matrix}
   ah+bg & a & h \\
   bf+ba & h & b \\
   af+bc & g & f \\
\end{matrix} \right|=\dfrac{a}{c}\left| \begin{matrix}
   ah+bg & a & hc \\
   bf+ba & h & bc \\
   af+bc & g & fc \\
\end{matrix} \right|(\text{c multiplied to 3rd column}) \\
 & =\dfrac{a}{c}\left| \begin{matrix}
   ah+bg & a & ab+hc \\
   bf+ba & h & bc+bh \\
   af+bc & g & bg+fc \\
\end{matrix} \right|\left( {{\text{c}}_{3}}\to {{\text{c}}_{3}}\text{+b}{{\text{c}}_{2}} \right)..........(1.4) \\
\end{align}$
Now, if we can prove that the difference of (1.3) and (1.4) is zero, we would prove the equation. Thus, defining the difference as $\delta $, we get
$\begin{align}
  & \delta =\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & gc & ab+ch \\
   bf+ba & fc & hb+bc \\
   af+bc & {{c}^{2}} & bg+fc \\
\end{matrix} \right|-\dfrac{a}{c}\left| \begin{matrix}
   ah+bg & a & ab+hc \\
   bf+ba & h & hb+bc \\
   af+bc & g & bg+fc \\
\end{matrix} \right| \\
 & =\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & gc & ab+ch \\
   bf+ba & fc & hb+bc \\
   af+bc & {{c}^{2}} & bg+fc \\
\end{matrix} \right|+\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & -{{a}^{2}} & ab+hc \\
   bf+ba & -ah & hb+bc \\
   af+bc & -ag & bg+fc \\
\end{matrix} \right|\text{(multiplying the overall factor a to column 2)} \\
 & \text{=}\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & gc-{{a}^{2}} & ab+ch \\
   bf+ba & fc-ah & hb+bc \\
   af+bc & c-ag & bg+fc \\
\end{matrix} \right|............(1.5) \\
\end{align}$
Now, we know that when two matrices are multiplied together, the corresponding rows and columns are multiplied in the matrix multiplication method. Also, the determinant of a product of matrices is equal to the product of the determinants, therefore (1.5) can be written as
$\begin{align}
  & \delta \text{=}\dfrac{1}{c}\left| \begin{matrix}
   ah+bg & gc-{{a}^{2}} & ab+ch \\
   bf+ba & fc-ah & hb+bc \\
   af+bc & c-ag & bg+fc \\
\end{matrix} \right| \\
 & =\dfrac{-1}{c}\left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|\times \left| \begin{matrix}
   0 & a & b \\
   -a & 0 & c \\
   -b & -c & 0 \\
\end{matrix} \right|............(1.6) \\
\end{align}$
Now, we know that the determinant of a general matrix is given by
$\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a(ei-fh)+b(fg-di)+c(dh-eg)........(1.7)$
Thus, using the expansion (1.7) in the second matrix of (1.6), we obtain
$\left| \begin{matrix}
   0 & a & b \\
   -a & 0 & c \\
   -b & -c & 0 \\
\end{matrix} \right|=0\left( 0+{{c}^{2}} \right)+a(-cb-0)+b(ac-0)=-abc+abc=0$
Using this in equation 1.6, we get
$\delta =\dfrac{-1}{c}\left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|\times \left| \begin{matrix}
   0 & a & b \\
   -a & 0 & c \\
   -b & -c & 0 \\
\end{matrix} \right|=\dfrac{-1}{c}\left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|\times 0=0$
Thus, we have proved that the difference between LHS and RHS ($\delta $) is 0. Therefore, LHS=RHS and thus we have proved the equation.

Note: We should note that to prove the equation, taking the difference and showing it to be zero was more useful that taking the LHS and simplifying it to make it equal to RHS. Therefore, in such questions, taking the difference of LHS and RHS and showing it to be zero is more useful.